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<h3 class="heading"><span class="type">Paragraph</span></h3>
<p>Back to Fourier series.For <span class="process-math">\(f(x)\)</span> periodic with period <span class="process-math">\(2L\text{,}\)</span> Fourier series for <span class="process-math">\(f(x)\)</span> is defined as:</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
\hat{f}(x)=\frac{a_0}{2}+\sum_{m=1}^{\infty}\left(a_m\cos\frac{m\pi x}{L} +b_m\sin\frac{m\pi x}{L}\right),\quad (*)
\end{equation*}
</div>
<p class="continuation">To determine <span class="process-math">\(a_i\)</span> and <span class="process-math">\(b_i\text{,}\)</span> we use the <dfn class="terminology">orthogonality</dfn> of the trig set!<dfn class="terminology">Steps:</dfn> Multiply <span class="process-math">\((*)\)</span> by <span class="process-math">\(\cos\frac{n\pi x}{L}\)</span> and integrate over <span class="process-math">\([-L, L]\text{.}\)</span></p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
\begin{aligned}\int_{-L}^L f(x)\cos\frac{n\pi x}{L} &amp;=&amp; \int_{-L}^L \frac{a_0}{2}\cos\frac{n\pi x}{L}+\sum_{m=1}^{\infty}\int_{-L}^L a_m\cos\frac{m\pi x}{L}\cos\frac{n\pi x}{L}\textrm{d}x\\
&amp; &amp; +\sum_{m=1}^{\infty}\int_{-L}^Lb_m\sin\frac{m\pi x}{L}\cos\frac{n\pi x}{L}\textrm{d}x.\end{aligned}
\end{equation*}
</div>
<p class="continuation">All the terms are <span class="process-math">\(0\)</span> except one:</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
\int_{-L}^L f(x)\cos\frac{n\pi x}{L} = \int_{-L}^L a_n\cos\frac{n\pi x}{L}\cos\frac{n\pi x}{L}\textrm{d}x=a_nL
\end{equation*}
</div>
<p class="continuation">This gives us the formula to compute <span class="process-math">\(a_n\text{:}\)</span></p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
{a_n=\frac{1}{L}\int_{-L}^L f(x)\cos\frac{n\pi x}{L}\textrm{d}x},\qquad n=1,2,3,\cdots
\end{equation*}
</div>
<p class="continuation">Similarly, multiplying <span class="process-math">\((*)\)</span> by <span class="process-math">\(\sin\frac{n\pi x}{L}\)</span> and integrating over <span class="process-math">\([-L, L]\text{,}\)</span> we derive the formula for <span class="process-math">\(b_n\text{:}\)</span></p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
{b_n=\frac{1}{L}\int_{-L}^L f(x)\sin\frac{n\pi x}{L}\textrm{d}x},\qquad n=1,2,3,\cdots
\end{equation*}
</div>
<p class="continuation">In a similar way, we get</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
{a_0=\frac{1}{L}\int_{-L}^L f(x)\textrm{d}x}.
\end{equation*}
</div>
<p class="continuation">Note that <span class="process-math">\(a_0/2\)</span> is the average of <span class="process-math">\(f(x)\)</span> over a period. The formula for <span class="process-math">\(a_0\)</span> fit in the one for <span class="process-math">\(a_n\)</span> with <span class="process-math">\(n = 0\text{.}\)</span> Hence we summarize the formulae in a more compacted way. These formulae for computing the Fourier coefficients are called <dfn class="terminology">Euler-Fourier formula</dfn>. If the period is <span class="process-math">\(2\pi\text{,}\)</span> i.e., <span class="process-math">\(2L = 2\pi\text{,}\)</span> we get simpler looking formulas</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
a_n=\frac{1}{\pi}\int_{-\pi}^{\pi} f(x)\cos nx\textrm{d}x,\qquad n=\textcolor{red}{0},1,2,3,\cdots
\end{equation*}
</div>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
b_n=\frac{1}{\pi}\int_{-\pi}^{\pi} f(x)\sin nx\textrm{d}x,\qquad n=1,2,3,\cdots.
\end{equation*}
</div>
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